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The three components of the curl cancel pairwise:The domain is all of ${\mathbb{R}}^3$, which is simply connected, so this is consistent with a conservative vector field. To construct its scalar potential, integrate ${\Phi }_x=y+z$ to obtain $\Phi =x(y+z)+h(y,z)$. Matching ${\Phi }_y=x+z$ gives $h_y=z$, hence $h=yz+k(z)$. Finally ${\Phi }_z=x+y$ forces $k^{\prime }(z)=0$. ConsequentlyEvery scalar potential differs by a constant on this connected domain. The endpoint values differ by one, agreeing with the line integral in part (a).

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.

For $\mathbf{r}(t)=(t^2,t{)}^T$, the tangent vector is $(2t,1{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,t^4+t{)}^T$. ThusThe unequal values for the same endpoints already rule out a conservative vector field.

On the curved path, $\mathbf{r}(t)=(t,t,t^2{)}^T$ and ${\mathbf{r}}^{\prime }=(1,1,2t{)}^T$. Direct substitution givesThusBoth values agree with part (i), for every $a,b,c$. Agreement on these two selected paths alone does not establish a conservative vector field.

On the circular arc, $z=0$ and $dz=0$, so the line integral reduces toThis exact differential evaluation applies to the specified arc; for example its parametric equation is $(x,y,z)=(1-\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving ${\int }_0^1(\lambda +1)dz=\lambda +1$. The whole path therefore givesTo answer the final conservativeness requirement, compute the curl:It vanishes identically exactly whenOn the simply connected whole space this condition is sufficient for a conservative vector field. Explicitly a vector-field potential isBoth path calculations then give $13/6$, equal to $\varphi (1,1,1)-\varphi (0,0,0)$ by the fundamental theorem for line integrals. Equality of just these two path values would only force $\mu =2\lambda +1$; it is the curl calculation that establishes full path independence.

The relevant mixed partial derivatives agree:The vector field is smooth on the entire simply connected plane, so this condition establishes that it is a conservative vector field. We can also construct its vector-field potential explicitly. From ${\varphi }_x=y^2-2x$, take an antiderivative in $x$:Then ${\varphi }_y=2xy+h^{\prime }(y)=2xy-y$ requires $h^{\prime }(y)=-y$. HenceThe fundamental theorem for line integrals checks the previous answer: $\varphi (1,1)-\varphi (0,0)=-1/2$.

A time-independent conservative force defines a conservative vector field on a domain when the work between two points is independent of path, equivalently when every closed-path work integral vanishes. Define a potential energy $U$ by $\boxed{\mathbf{F}=-\nabla U}$ and $U(B)-U(A)=-{\int }_A^B\mathbf{F}\cdot d\mathbf{r}$. The additive constant is arbitrary; only differences affect work. A vanishing curl is necessary, and is sufficient on a simply connected domain under the usual smoothness assumptions. Local zero curl by itself need not guarantee a single-valued global potential on a domain with holes.

For a test mass outside a point or spherical mass, $U=-GMm/r$ with zero at infinity reproduces the inward gravitational force. Near a surface with approximately uniform downward gravity, $U=mgz$ when $z$ increases upward. A linear spring has $U=k{x}^2/2$. In electrostatics the electric potential is energy per unit charge, so $U=qV$ and $\mathbf{E}=-\nabla V$; potential and potential energy have different units and roles.

For motion under a time-independent conservative force, $d(K+U)/dt=0$. nonconservative forces change the mechanical energy by their work, and an explicitly time-dependent potential can exchange energy with the external agency changing it. A time-varying magnetic field can give a circulating electric field, so not every electric field is electrostatic or conservative. Conversely, zero magnetic work on a charge does not turn a velocity-dependent magnetic force into an ordinary position-potential force.

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

A line integral is path dependent when two admissible paths with the same starting and ending points give different values. This disproves path independence and, for a vector field, shows that it is not a conservative vector field on the relevant domain.

A line integral is path independent when its value depends only on the two endpoints. On a connected domain, this holds for the line integral of every exact differential.

For a vector-field line integral, this condition is equivalent to the field being a conservative vector field.

A potential for a conservative vector field $X$ is a function with scalar values $\Phi$ whose gradient is $X$. Potentials differ by a constant on each connected component. The fundamental theorem for line integrals gives ${\int }_{\gamma }X\cdot d\mathbf{x}=\Phi (\gamma (1))-\Phi (\gamma (0))$. In mechanics a conservative force is conventionally written $-\nabla V$, so its potential energy is $V=-\Phi$. This spatial potential differs from the scalar potential $V(\varphi )$ that denotes the derivative-free energy density of a scalar field.